Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $t = \dfrac{-5k - 40}{-k^2 + 4k + 45} \times \dfrac{k^3 + 3k^2 - 10k}{-k^2 + 2k} $
First factor out any common factors. $t = \dfrac{-5(k + 8)}{-(k^2 - 4k - 45)} \times \dfrac{k(k^2 + 3k - 10)}{-k(k - 2)} $ Then factor the quadratic expressions. $t = \dfrac {-5(k + 8)} {-(k + 5)(k - 9)} \times \dfrac {k(k + 5)(k - 2)} {-k(k - 2)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {-5(k + 8) \times k(k + 5)(k - 2) } { -(k + 5)(k - 9) \times -k(k - 2)} $ $t = \dfrac {-5k(k + 5)(k - 2)(k + 8)} {k(k + 5)(k - 9)(k - 2)} $ Notice that $(k + 5)$ and $(k - 2)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {-5k\cancel{(k + 5)}(k - 2)(k + 8)} {k\cancel{(k + 5)}(k - 9)(k - 2)} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $t = \dfrac {-5k\cancel{(k + 5)}\cancel{(k - 2)}(k + 8)} {k\cancel{(k + 5)}(k - 9)\cancel{(k - 2)}} $ We are dividing by $k - 2$ , so $k - 2 \neq 0$ Therefore, $k \neq 2$ $t = \dfrac {-5k(k + 8)} {k(k - 9)} $ $ t = \dfrac{-5(k + 8)}{k - 9}; k \neq -5; k \neq 2 $